Given: The systolic arterial blood pressure observed for 20 dogs is normally distributed with a mean of 152 mm of mercury (Hg) and a standard deviation of 18 mm of Hg.
To find: P(100 < 152)
Method: Calculation of Z-Score followed by the probability or area of the bell curve at X = 100.
Solution:
Mean u = 152, std s = 18
Z score = 
The value of P(100<152) is calculated by looking at the value of Z in the Z score for the standard normal distribution given in the image.
P(Z=-2.89) = 0.0019
The P(Z = -2.89) corresponds to the area in the left tail of the bell curve.
Thus the probability of 100 mm Hg blood pressure is 0.0019.
I have provided the options below:
a. 150
b. 225
c. 300
d. 450
e. 75
Answer:
150
Explanation:
Hardy-Weinberg law states, the allelic frequencies of a gene that is, q and p follow the relationship p^2+q^2+2pq = 1, if the population is in equilibrium.
Therefore p^2+q^2+2pq = 1, can be used to calculate the percentage of the allelic frequencies i.e. p^2 and q^2
Allelic frequency = 0.75.
Therefore, the percentage of the allele in the population would be given by
0.75^2 = 0.5625 = 56%.
And the total number of individuals with this allele in the population:
number of individuals = 56×300÷100
= 168 or
150 nearest figure.
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Explanation:
It is based on the principle of continuous improvement. The standard enables companies to define what a quality product should be and how it should meet customer needs.
If a student rests and then squeezes a clothespin for one minute, will they be able to squeeze it more than someone that exercised before squeezing the pin for 1 minute?
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